Sirindhorn International Institute of Technology Thammasat University at Rangsit School of Information, Computer and Communication Technology ECS33: Midterm Examination (Set I) COURSE : ECS33 (Principles of Communications) DATE : August 1, 011 SEMESTER : 1/011 INSTRUCTOR : Dr.Prapun Suksompong TIME : 13:30-16:30 PLACE : BKD 3507 Name ID Seat Instructions: 1. Including this cover page, this exam has 1 pages.. Read these instructions and the questions carefully. 3. Fill out the form above. 4. Closed book. Closed notes. 5. Basic calculators, e.g. FX-991MS, are permitted, but borrowing is not allowed. 6. Unless specified otherwise, write down all the steps that you have done to obtain your answers. You may not get any credit even when your final answer is correct without showing how you get your answer. 7. Allocate your time wisely. 8. Do not forget to write your first name and the last three digits of your ID in the spaces provided on the top of each examination page, starting from page. 9. Allocate your time wisely. 10. When not explicitly stated/defined, all notations and definitions follow ones given in lecture. 11. Some points are reserved for reducing answers into their simplest forms. 1. Do not cheat. The use of communication devices including mobile phones is prohibited in the examination room. 13. Do not panic. Page 1 of 1
1. (5 pt) For the signal gt shown below. Sketch g4 t needed.. No explanation is 8 gt -3 5 t. (5 pt) The Fourier transform X f for a signal Let yt x4 t. Sketch xt is shown below. Y f. No explanation is needed. 8 X f -3 5 f 3. (5 pt) The Fourier transform of the Gaussian pulse G f e f Fourier transform of is given by g t e t. Using this information and the time-scaling property, find the. x t e 3t Page of 1
4. (5 pt) Suppose g t 1 t a. ( pt) Find G f.. b. ( pt) Find the smallest positive zero-crossing of G f. c. (1 pt) A non-accurate plot of G f is given below. Your answer in part (b) is denoted by f 1. The next smallest positive zero-crossing of G f is denoted by f 3. Between f 1 and f 3, there is a local minimum at f. f1 f3 Compare f and which one is larger?. Theoretically, should they be the same? If not, f f f 1 f 3 5. (5 pt) In this question, you are provided with a partial proof of an important result in the study of Fourier transform. Your task is to figure out the quantities/expressions inside the boxes labeled a,b,c, and d. Put your answers in the spaces provided at the end of the question. No explanation is needed. We start with a function that involves expression for gt. Then, we define xt g t T xt in terms of a sum that involves g t. What you will see next is our attempt to find another G f.. It is a sum Page 3 of 1
To do this, we first write xt as xt g t* t T convolution-in-time property, we know that X We can get f is given by X f G f a f b. xt back from X. Then, by the f by the inverse Fourier transform formula: j ft. After plugging in the expression for X x t X f e df we get j ft x t e G f a f b df j ft a e G f f b df By interchanging the order of summation and integration, we have j ft x t a e G f f b df. f from above, We can now evaluate the integral via the sifting property of the delta function and get c x t a e G d.. a = c = b = d = 6. (10 pt) You are given the baseband signal mt 3cos 500t cos 1000t a. (5 pt) Sketch the spectrum of mt. No explanation is needed.. Page 4 of 1
b. (5 pt) Sketch the spectrum of the DSB-SC signal mtcos 5,000 t explanation is needed.. No 7. (8 pt) Suppose mt M f is bandlimited to W, i.e., M f 0 for f W. Consider the following DSB-SC transceiver. Transmitter (modulator) mt xt Delayed by cos c t xt FWR vt H f LP yt Receiver (demodulator) Also assume that c f W and that H f LP 1, f W 0, otherwise. Make an extra assumption that mt 0 for all time t and that the full-wave rectifier (FWR) input-output relation is described by a function f FWR : f FWR (Questions start on the next page.) x x, x 0, x, x 0. Page 5 of 1
a. Recall that the half-wave rectifier input-output relation is described by a x, x 0, function f HWR : fhwr x 0, x 0. We have seen in class and in HW that when the receiver uses half-wave rectifier, v t x t g t, HWR where g t 1cos t 0 HWR c. i. (3 pt) The receiver in this question uses full-wave rectifier. Its vt can be described in a similar manner; that is v t x t g t. Find gfwr t. Hint: FWR g t c g t c for some constants c 1 and c. Can 1 HWR you find these constants? FWR ii. ( pt) Recall that the Fourier series expansion of ghwr t is given by 1 1 1 1 ghwr t cosct cos3 ct cos5 ct cos7 ct 3 5 7. Find the Fourier series expansion of g t. FWR Page 6 of 1
b. (3 pt) Find yt (the output of the LPF). 8. (1 pt) The signal xt cos t at 3 seconds. x( t) : cos t at 3 1 0.5 is plotted below. The time unit is in x( t) 0 0.5 1 0 1 3 4 1 0.5 t x( t) 0 0.5 1 1.9.1 t Page 7 of 1
a. ( pt) Find the instantaneous frequency (in Hz) at t. Hint: it is an integer. b. (5 pt) Find the value of a. Hint: it is an integer. c. (5 pt) Find the instantaneous frequency (in Hz) at t 5. Hint: it is an integer. 9. (10 pt) A modulated signal with carrier frequency equation 10cos c 5sin 3000 a. (5 pt) Find the instantaneous frequency f t of x t f t t. 5 f 10 Hz is described by the c xt. b. (5 pt) Estimate the bandwidth of xt via Carson s rule. Page 8 of 1
10. (8 pt) A modulated signal with carrier frequency equation 5 f 10 Hz is described by the xt 10cos fct 5sin 3000t 10sin 000t a. (5 pt) Find the instantaneous frequency f t of xt.. c b. (3 pt) Estimate the bandwidth of xt via Carson s rule. 11. (16 pt) Determine the Nyquist sampling rate and for the signals in the table below. No explanation is needed. 00 t t t t t sinc 00 t sinc sinc 00 5sinc 10 sinc 100 sinc 00 Nyquist sampling rate Nyquist sampling interval Page 9 of 1
1. (5 pt) Let ct A, 0t T 0, otherwise You may assume that A > 0. A message m m, m, m,, m 0 1 9 (-1 1 1-1 -1-1 1-1 1-1) is sent via 1 k where is the length of m. k0 s t m c t kt The magnitude S f of the spectrum is plotted below. 70 60 50 40 30 0 10 0-0.5-0.4-0.3-0. -0.1 0 0.1 0. 0.3 0.4 0.5 f [khz] a. ( pt) Find T. b. (3 pt) Find A. Page 10 of 1
13. (6 pt) Use properties of Fourier transform to evaluate the following integrals. sin x (Recall that sinc x.) Clearly state which properties that you use. x a. (3 pt) sinc 5x dx * f j f 5 j b. (1 pt) sinc sinc e f e f df c. (1 pt) sinc 5xsinc 7 x dx Page 11 of 1
7 sinc x 5 sincx dx d. (1 pt) Extra Credits: Complete the following lyrics of the Fourier's Song Integrate your function times a complex It's really not so hard you can do it with your pencil And when you're done with this calculation You've got a brand new function - the Fourier Transformation What a prism does to sunlight, what the ear does to sound Fourier does to signals, it's the coolest trick around Now filtering is easy, you don't need to All you do is multiply in order to solve. From time into frequency - from frequency to time Every operation in the time domain Has a Fourier analog - that's what I claim Think of a delay, a simple in time It becomes a phase rotation - now that's truly sublime! And to differentiate, here's a simple trick Just multiply by j omega, ain't that slick? Integration is the inverse, what you gonna do? Divide instead of multiply - you can do it too. cos x1cos sin x1cos j ft G f g t e dt cos 1 1 fct f fc e f fc e j j ft0 0 j f0t g t t e G f e g t G f f x x 0 1 1 g tcos fct G f fc G f fc sin x 1t a a sinc fa where sinc x x j Page 1 of 1